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这个问题似乎并不涉及在help Center中界定的范围之内的方案规划。

Closed 3 hours ago.

我没有像现在这样面临两难行动的问题:

Generate mask indicating leftmost 1 in x. Assume w=32.
For example, 0xFF00 -> 0x8000, and 0x6600 -> 0x4000.
If x = 0, then return 0.

The question also sets the limits that only using bitwise operations and "-" "+" to solve the problem. Here s detail:

隐蔽

Conditionals (if or ?:), loops, switch statements, function calls, and macro invocations. Division, modulus, and multiplication. Relative comparison operators (<, >, <=, and >=).

允许的行动

All bit-level and logic operations. Left and right shifts, but only with shift amounts between 0 and w – 1. Addition and subtraction. Equality (==) and inequality (!=) tests. (Some of the problems do not allow these.) Integer constants INT_MIN and INT_MAX. Casting between data types int and unsigned, either explicitly or implicitly.”

我可以理解这些问题的解决办法。但是,我第一次没有接触过。这是否意味着我缺乏任何知识,例如离散的数学? 是否有任何来源来开展高效的双向行动?

Answer 1

在这里,我们采取了一种冷静的做法:

Initialize a 32 bit unsigned integer, m, with only the highest bit set (1u << 31).
Loop until m reaches 0, right shifting it one step at the end of every iteration.
In the loop, check if m & value_to_check (bitwise AND) is non-zero. If it is, break out.
After the loop, m will have the leftmost bit in value_to_check set (or be zero if value_to_check was zero).

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